2006-11-07 01:31:47 · 4 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
m^6-n^6=(m^3+n^3)(m^3-n^3)
(m^2-n^2)(m^4+m^2n^2+n^4)
m^4-n^4=(m^2+n^2)(m^2-n^2)
plugging in
m^6-n^6/m^4-n^4
=(m^4+m^2n^2+n^4)/(m^2+n^2)
2006-11-07 01:38:48 · answer #1 · answered by raj 7 · 0⤊ 1⤋
(m^3-n^3)(m^3+n^3) over (m-n)(m-n)(m^2+n^2)
2006-11-07 01:47:04 · answer #2 · answered by Anonymous · 0⤊ 0⤋
m^6-n^6=(m^3-n^3)(m^3+n^3)
=(m-n)(m^2+mn+n^2)(m^3+n^3) --------(1)
m^4-n^4=(m^2-n^2)(m^2+n^2)
=(m+n)(m-n)(m^2+n^2) --------(2)
(1)/(2)=(m^2+mn+n^2)(m^3+n^3) /(m^2+n^2) (m+n)
2006-11-07 01:53:41 · answer #3 · answered by Naveen 2 · 0⤊ 0⤋
(m^6 - n^6)/(m^4-n^4)
=[(m^3-n^3)(m^3+n^3)]/[m^2+n^2)(m^2-n^2)]
=[(m-n)(m^2+mn+n^2)(m^3+n^3)]/[m^2+n^2)(m+n)(m-n)]
2006-11-07 01:53:25 · answer #4 · answered by ironduke8159 7 · 0⤊ 0⤋